In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used instead for alpha particles with energy 300 keV . An alpha particle has charge q=+2e and massm=6.64×10−27kg .If the magnetic field isnt changed, what will be the orbital radius of the alpha particles? Express your answer to three significant figures and include the appropriate units.
Answers
Answer:
16 cm
Explanation:
For protons:
Energy, E = 300 keV
radius of orbit, r1 = 16 cm
the relation for the energy and velocity is given by
So, .... (1)
Now,
Substitute the value of v from equation (1), we get
Let the radius of the alpha particle is r2.
For proton
So, ... (2)
Where, m1 is the mass of proton, q1 is the charge of proton
For alpha particle
So, ... (3)
Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle
Divide equation (2) by equation (3), we get
q1 = q
q2 = 2q
m1 = m
m2 = 4m
By substituting the values
So, r2 = r1 = 16 cm
Thus, the radius of the alpha particle is 16 cm.
Answer:15.95 cm
Explanation:
Given
Energy=300 kev
radius of Proton=16 cm
mass of alpha particle
mass of proton
charge on alpha particle is twice of proton
radius of Proton is given by
and Kinetic energy
where P=momentum
---1
Radius for Alpha particle is
-----2
Divide 1 & 2 we get
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If a person’s weight is W on the surface of the earth, calculate what it would be, in terms of W, at the surface of (a) the moon; (b) Mars; (c) Jupiter.
Answers
Answer:
3 fans per 15 A circuit
Explanation:
From the question and the data given, the light load let fan would have been
(60 * 4)/120 = 240/120 = 2 A.
Next, we add the current of the fan motor to it, so,
2 A + 1.8 A = 3.8 A.
Since the devices are continuos duty and the circuit current must be limited to 80%, then the Breaker load max would be
0.8 * 15 A = 12 A.
Now, we can get the number if fans, which will be
12 A/ 3.8 A = 3.16 fans, or approximately, 3 fans per 15 A circuit.
Final answer:
The total power draw of each fan is 3.8 amperes. Thus, considering a limit of 80% usage of 15 amperes, only 3 fans can be connected to a single circuit to keep the total power draw below 12 amperes.
Explanation:
The question is asking how many ceiling fans, each with a certain power draw, can be connected on a single 15-ampere circuit, considering that each fan is a continuous-duty device. The power draw of each fan when the motor is operated at high speed and the light kit is fully loaded is the sum of the power draw of the motor and the light kit. As the power draw of each motor is 1.8 amperes and the light kit is 240 watts or 2 amperes (calculated using the formula Power = Voltage x Current; assuming a voltage of 120 volts), the total power draw of each fan is 3.8 amperes. Considering the limit of 80% of the continuous load, only 12 amperes (80% of 15) can be used. Thus, 3 fans can be connected to the circuit as it reaches 11.4 amperes, close enough to the 12 amperes limit.
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Answers
Answer:
Explanation:
Given data
Electric Field E=2.5×10⁵ N/C
To find
Charge Density
Solution
From definition of charge density we know that:
Charge Density=Electric field×Permttivity
Where Permttivity ∈₀=8.85×10⁻¹²C²/N.m²
Answers
Final answer:
The torque about a given origin when a force N is acting on a particle at the position vector m is given by the cross product of the position and force vectors. It's represented by the SI unit Newton-meters, and for multiple particles, the total angular momentum is the vector sum of their individual angular momenta.
Explanation:
The torque about a given origin, when a force N is acting on a particle located at the position vector m, is calculated using the cross product of the position vector and the force vector. This can be written as τ = m x N. The SI unit of torque is Newton-meters (N.m).
As an example, if you apply a force perpendicularly at a distance from a pivot point, you will create a torque relative to that point. Similarly, the torque on a particle is also equal to the moment of inertia about the rotation axis times the angular acceleration.
If we consider multiple particles, the total angular momentum of these particles about the origin is the vector sum of their individual angular momenta. This is calculated by the expression for the angular momentum Ỉ = ŕ x p for each particle, where ŕ is the vector from the origin to the particle and p is the particle's linear momentum.
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Final answer:
The torque on a particle at a position vector m with force N acting on it is calculated by taking the cross-product of the position vector and the force. This principle is the same even in systems with multiple particles. The SI unit of torque is Newton-meters (N·m), which should not be confused with Joules (J).
Explanation:
The torque on a particle located at a position vector m with a force N acting on it is calculated by taking the cross-product of the position vector and the force. In terms of physics, torque (τ) is a measure of the force that can cause an object to rotate about an axis, and it is calculated as the product of the force and the distance from the axis of rotation to the point where force is applied. Hence, the formula for torque is τ = r x F where r is the position vector (or distance from the origin to the point where the force is applied) and F is the force. Remember, this equation gives a vector result with a direction perpendicular to the plane formed by r and F and a magnitude equal to the product of the magnitudes of r and F and the sine of the angle between r and F.
The same principle applies to systems where multiple particles are present. The total angular momentum of the system of particles about a particular point is the vector sum of the individual angular momenta about that point. Torque is the time derivative of angular momentum.
The SI unit for torque is Newton-meters (N·m), which should not be confused with Joules (J), as both have the same base units but represent different physical concepts. In this context, a net force of 40N acting at a distance of 0.800m from the origin would generate a torque of 32 N·m at the origin.
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Answers
Answer:
-847.2J
Explanation:
First find the acceleration from v^2= u^2 + 2as
v= 2.5 m/s
u= 1.3 m/s
a???
s=6.00
a= v^2-u^2/2s
a= (2.5)^2-(1.3)^2/2× 6
a= 0.38ms^-2
From Newtons second law:
(Force applied cos Θ) - (Frictional force) = ma
Frictional force = ma- (Force applied cos Θ)
Frictional force= (18.8×0.38) - (165 cos 26°)
Frictional force= 7.144- 148.3= -141.2N
Therefore,
Work done by friction = Frictional force × distance covered
= -141.2N × 6= -847.2J
Answer:
W = –847J
Explanation:
Given m = 18.8kg, F = 165N, θ = -26° (below the horizontal, s = 6.0m, u = 1.30m/s and v = 2.50m/s
In this problem, two forces act on the chair; the forward force F and the frictional force f. We would apply newton's second law to find the frictional force f after which we can calculate the workdone by the frictional force f×s.
But for us to apply newton's second law, we need to know the acceleration of the chair cause by the net force.
From constant acceleration motion equations
v² = u² + 2as
2.5² = 1.30² + 2a×6
6.25 = 1.69 +12a
12a = 6.25 – 1.69
12a = 4.56a
a = 4.56/12
a = 0.38m/s
By newton's second law the net sum of forces equals m×a
The force F has horizontal and vertical and components. It is the horizontal component of this force that pushes the chair against friction.
Fx and f are oppositely directed.
So
Fx – f =ma
165cos(-26) – f = 18.8×0.38
148.3 – f = 7.14
f = 148.3 – 7.14
f = 141.2N
Workdone = -fs = –141.2×6.00 = –847J
W = –847J
Work is negative because it is done by a force acting on the chair in a direction opposite (antiparallel) to that of the intended motion.
A. Hydrosphere
B Lithosphere
C. Atmosphere
2
A Lithosphere
B Atmosphere
C Biosphere
Answers
2. B Atmosphere
lithosphere and biosphere
srry ik im super late but to help other people theres the answer
Answers
Answer:
The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.
Explanation:
Hi there!
The equation of height and velocity of the package are the following:
h = h0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
h = height of the package at time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).
v = velocity of the package at a time t.
First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:
v = v0 + g · t
0 = 50.5 m/s - 9.8 m/s² · t
Solving for t:
-50.5 m/s / -9.81 m/s² = t
t = 5.15 s
Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:
h = h0 + v0 · t + 1/2 · g · t²
h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²
h = 140 m
The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.